compute $\sum_{j=0}^{m} 3^j {m \choose j}$. Then, use the binomial theorem to verify the result.
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Please edit the title to make it more informative. – Frenzy Li Oct 29 '13 at 01:53
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how can I edit it? – math Lee Oct 29 '13 at 01:58
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To edit, click the "edit" button under your question. Make it informative, so that people seeing the title alone, have a sense of what you're conveying in the main body. – Frenzy Li Oct 29 '13 at 01:59
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thanks, I put my problem in the title. – math Lee Oct 29 '13 at 02:04
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$$4^m=(3+1)^m=\sum_{j=0}^m{m\choose j}\cdot3^j\cdot1^{m-j}=\sum_{j=0}^m{m\choose j}\cdot3^j$$ or $$\sum_{j=0}^m{m\choose j}\cdot3^j=\sum_{j=0}^m{m\choose j}\cdot3^j\cdot1^{m-j}=(3+1)^m=4^m$$
Lucian
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can you make it more clear, I still don't get why is 4^m in there and what does that C mean?' – math Lee Oct 29 '13 at 02:00
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1OP wanted to compute it and then "use binomial theorem to verify". So shouldn't some other method be used also? – Pratyush Sarkar Oct 29 '13 at 02:02
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@mathLee It is obvious from context that $C^j_m = \binom{m}{j}$. – Pratyush Sarkar Oct 29 '13 at 02:03
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1So $\sum_{j=0}^{m} 3^j {m \choose j}$ = 4^m finally? It is a backward computation? – math Lee Oct 29 '13 at 02:05
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$(a+b)^m=\sum_{j=0}^m{m\choose j}\cdot a^j\cdot b^{m-j}$ . See binomial theorem. – Lucian Oct 29 '13 at 02:13